Two cyclists left simultaneously from cities a and b heading towards each other at constant rates and met in 5 hours. the rate of the cyclist from a was 3 mph less than the rate of the other cyclist. if the cyclist from b had started moving 30 minutes later than the other cyclist, then the two cyclists would have met 31.8 miles away from a. what is the distance between a and b, in miles?

73.92 miles

Step-by-step explanation:

Formula for calculating distance = speed × time

Rate/speed of cyclist from city A = ( x - 3 ) mph , Time Taken = 5 Hours

Rate/speed of cyclist from city B = x mph

If Cyclist from B had started moving 30 minutes later

Time Taken by cyclist from B = 4.5 Hours

If B had started 30 minutes later, then the two cyclist would have met 31.8 miles from A.

distance covered by cyclist from A = 31.8

Solution:

For city A:

distance = speed × time                                                

31.8  = ( x - 3 ) × 5     ⇒   = x - 3

6.36 = x - 3    ⇒   x = 6.36 + 3   ⇒    x = 9.36 mph

For city B:  

distance =  speed × time

              =  9.36 × 4.5

              =  42.12 miles

Distance between A and B = 31.8 + 42.12

                                             = 73.92 miles

-2x + 7 = 7

subtract 7 from both sides.

-2x = 0

divide both sides by -2.

x = 0